If 3^(2sin2A - 1), 14, 3^(4 - 2sin2A) are 3 terms in an A.P, find the 5th term.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that the middle term is the arithmetical mean of the the terms `3^(2sin2A - 1)`  and `3^(4 - 2sin2A)`  such that:

`14 = (3^(2sin2A - 1) + 3^(4 - 2sin2A))/2`

You need to use the properties of exponentials such that:

`3^(2sin2A - 1) = (3^(2sin2A))/3`

`3^(4 - 2sin2A) = 3^4 /(3^(2sin2A))`

`28 = (3^(2sin2A))/3 + 3^4 /(3^(2sin2A))`

`84*3^(2sin2A) = (3^(2sin2A))^2 + 3^5`

You should come up with the substitution `3^(2sin2A) = y`  such that:

`84y = y^2 + 243`

`y^2 - 84y+ 243 = 0`

`y_(1,2) = (84 +- sqrt(84^2 - 4*243))/2`

`y_(1,2) = (84 +- sqrt(6084))/2`

`y_(1,2) = (84 +-78)/2`

`y_1 = 81 ; y_2 = 3`

You need to solve for sin 2A the equations  `3^(2sin2A) = 81`  and `3^(2sin2A) = 3`  such that:

`3^(2sin2A) = 3^4 =gt 2 sin 2A = 4`

`sin 2A = 2`  impossible since the sine of an angle is not larger than 1

`3^(2sin2A) = 3 =gt 2 sin 2A = 1 =gt sin 2A = 1/2`

You need to evaluate the terms `3^(2sin2A - 1)`  and `3^(4 - 2sin2A)`  substituting `1/2 ` for `sin 2A`  such that:

`3^(2sin2A - 1) = 3^(2*(1/2) - 1) = 3^0 = 1`

`3^(4 - 2/2) = 3^3 = 27`

You need to find the common difference such that:

`d = 27 - 14 = 13`

`d = 14 - 1 = 13`

You need to find the 4th term such that:

`a_4 = 27 + 13 = 40`

`a_5 = 40 + 13 = 53`

Hence, evaluating the 5th term of the arithmetical progression under given conditions yields  `a_5 = 53.`

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